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Questions about mesh initialization in CONVERGE |
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April 5, 2019, 16:05 |
Questions about mesh initialization in CONVERGE
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#1 |
Member
Mingyuan Tao
Join Date: Mar 2016
Posts: 31
Rep Power: 10 |
Hi all,
I have some questions about the way CONVERGE use to initialize mesh. I have a rectangular geometry with L=0.2m, and H=0.05m, and I am conducting mesh independent calculation currently. However, there are two problems: 1. I have the fix_embedding on with 20 layers and scales=2. If I set dx=6e-4, dy=8e-4, grid_scale=0, it gives a basically good fix_embedding on the boundary. However, if I set dx=1.2e-3, dy=1.6e-3, grid_scale=1, which in my understanding, should function the same as the previous setting, the generated mesh shows a much thinker fix embedding on the boundary, and therefore more cells are generated for calculation. I am really confused about the difference between these two settings. 2. Another confusing problem also arises from the two different dx, dy, grid_scale settings. When I use parallel calculation command: "nohup mpirun -n 50 converge-2.4.man super >a.log &" the memory usage by setting dx=6e-4, dy=8e-4, grid_scale=0 is much higher than the usage by setting dx=1.2e-3, dy=1.6e-3, grid_scale=1. Sometimes depending on how small dx, dy can be, the case with grid_scale=0 cannot even be calculated because almost all memory is used up. Could anyone help me these two problems? I appreciate it! |
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April 5, 2019, 19:23 |
Re: Questions about mesh initialization in CONVERGE
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#2 |
New Member
Vigneshwar Ravisankar
Join Date: Apr 2019
Posts: 26
Rep Power: 7 |
Hi,
1. The embedded grid are generated based on the base grid value at each scale and in further calculation the grid scaling is performed. So in case of dx=6e-4, dy=8e-4, grid_scale=0, the base grid is considered as dx=6e-4, dy=8e-4. And further embedding is performed for scaling of 2. In case of dx=1.2e-3, dy=1.6e-3, grid_scale=1, The dx and dy at scale=0 is dx=1.2e-3, dy=1.6e-3. So this is used as base grid for initial embedding. Then for further calc, the base grid is re-scaled to dx=6e-4, dy=8e-4 and embedding is done to scale 2 with new base grid value. 2. The block sice for parallel calculation is calculated as: Block size= dx * 2^PS, Where PS is parallel scaling. Assuming PS is 1. So for the first case, Block size is 1.2e-3 (=6e-4*2), 1.6e-3 (=8e-4*2). so for a domain of L=0.2m, and H=0.05m. # block will be close to 5208, with 4 grids in each. (roughly in simple terms) For the 2nd case Block size is 2.4e-3 (=1.2e-3*2), 3.2e-3 (=1.6e-3*2) so for a domain of L=0.2m, and H=0.05m. # block will be close to 1302 (1/4th), with 4 grids in each. (roughly in simple terms). So the first case using a lot more memory due to high # of blocks. Change the PS value accordingly for better load balance Hope that's clear. Let me know if you still have trouble to understand my explanation. Best regards, __________________ Vigneshwar Ravisankar Application Engineer CONVERGECFD |
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April 5, 2019, 20:19 |
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#3 | |
Member
Mingyuan Tao
Join Date: Mar 2016
Posts: 31
Rep Power: 10 |
Thanks for your reply, that's clear!
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